# Intersection of Line Segments (1 Viewer)

#### zacn

##### New Member
Consider pair of line segments:
2i + 3j + lambda(-2i + 3j)
and
6i + 2j + lambda(i - 4j)

What coordinate is the intersection of both segments?

#### ultra908

##### Active Member
firstly, these are vector equations of lines, not line segments, unless the domain of lambda is restricted between some finite values.
Also, not that the lambda in these two lines aren't really the same variable- they can take on different values. It might be easier to notate one of them as mu or something else. (though in questions they might give them both to you as lambda)

When the lines intersect, 2i+3j +lambda(-2i+3j) = 6i +2j + mu(i-4j).
Equating i and j components will give you a set of simultaneous equations to solve for lambda and mu. Then subbing the value of lambda (or mu) into the appropriate line equation will give you the position vector of the point of intersection

#### Drongoski

##### Well-Known Member
Consider pair of line segments:
2i + 3j + lambda(-2i + 3j)
and
6i + 2j + lambda(i - 4j)

What coordinate is the intersection of both segments?
As pointed out by ultra908, question asks for the point of intersection of 2 lines:

$\bg_white L1: \binom x y = \binom 2 3 + \lambda \binom {-2} 3 \\ \\ L2: \binom x y = \binom 6 2 + \mu \binom 1 {-4}$

Rewriting these vector equations in their parametric form:

$\bg_white L1: x = 2-2\lambda and y = 3 + 3\lambda\\ \\ L2: x = 6 + \mu and y = 2 - 4\mu$

Equating x and y for the 2 lines, we get after simplifying:

$\bg_white 2\lambda + \mu = -4 and 3\lambda + 4\mu = -1 and solving, we get \lambda = -3$

Using this, we get: x = 2 -2(-3) = 8 and y = 3 + 3(-3) = -6

So the point of intersection is (8, -6)

The line L1 passes thru the point (2,3) with gradient 3/(-2) = -3/2 with eqn: 3x + 2y = 12

and L2 passes thru the point (6,2) with gradient (-4)/1 = -4 with eqn: 4x + y = 26

The point (8 ,-6) satisfies both these equations; therefore (8, -6) lies on both lines and is therefore the point of intersection of L1 and L2. This also indirectly gives me the satisfaction that I had got the value of the parameter $\bg_white \lambda = -3$ correct.

Remark

One way to specify a line is to provide one point on the line and its direction. In typical co-ord geometry this direction is the gradient.

In vector method, the direction is provided by the direction vector (not unique), e.g. for L1 it is $\bg_white \binom {-2} 3$

In L!: the vector eqn is based on the one point on L1, (2,3) and the said direction vector.

Last edited:

#### beetree1

##### Well-Known Member
Unrelated but does anyone know if internal and external dividing is still in the syllabus for 3u and or 4u

#### ultra908

##### Active Member
Unrelated but does anyone know if internal and external dividing is still in the syllabus for 3u and or 4u
yes using vectors